Question

Sketch the graph of the function. Choose a scale that allows all relative extrema and points of inflection to be identified on the graph. \(y=x^{3}-6 x^{2}+3 x+10\)

Short answer

The critical points are determined by solving \(3x^{2}-12x+3=0\) and the points of inflection by solving \(6x-12=0\). With these points along with the origin and y-intercept marked, one can sketch the function.

Step-by-step solution

Find the derivative

First find the derivative of the given function. The derivative of \(y=x^{3}-6x^{2}+3x+10\) is \(y'=3x^{2}-12x+3\).

Set the derivative equal to zero

To find the critical points, set the derivative equal to zero, i.e solve the equation \(3x^{2}-12x+3=0\).

Calculate the second derivative

To find the points of inflection and relative extrema, calculate the second derivative. The second derivative of the given function is \(y''=6x-12\).

Set the second derivative equal to zero

Set the second derivative equal to zero, and solve for \(x\) to find the points of inflection. ie. \(6x-12=0\)

Sketch the graph

Now it's time to sketch the graph. On the y-axis, mark the points where the function crosses it (origin and y-intercept). Mark the critical points and the points of inflection on the x-axis. Also mark any relative extrema, marking whether they're local maximums or minimums. You have now marked all the necessary points which will help you sketch your graph.